Chapter 10

CHAPTER 10
STABILITY OF SLOPES
10.1 INTRODUCTION
Slopes of earth are of two types
1. Natural slopes
2. Man made slopes
Natural slopes are those that exist in nature and are formed by natural causes. Such slopes
exist in hilly areas. The sides of cuttings, the slopes of embankments constructed for roads,railway
lines, canals etc. and the slopes of earth dams constructed for storing water are examples of man
made slopes. The slopes whether natural or artificial may be
1. Infinite slopes
2. Finite slopes
The term infinite slope is used to designate a constant slope of infinite extent. The long slope
of the face of a mountain is an example of this type, whereas finite slopes are limited in extent. The
slopes of embankments and earth dams are examples of finite slopes. The slope length depends on
the height of the dam or embankment.
Slope Stability: Slope stability is an extremely important consideration in the design and
construction of earth dams. The stability of a natural slope is also important. The results of a slope
failure can often be catastrophic, involving the loss of considerable property and many lives.
Causes of Failure of Slopes: The important factors that cause instability in a slope and lead to
failure are
1. Gravitational force
2. Force due to seepage water
3. Erosion of the surface of slopes due to flowing water
365

366 Chapter 10
4. The sudden lowering of water adjacent to a slope
5. Forces due to earthquakes
The effect of all the forces listed above is to cause movement of soil from high points
to low points. The most important of such forces is the component of gravity that acts in the
direction of probable motion. The various effects of flowing or seeping water are generally
recognized as very important in stability problems, but often these effects have not been
properly identified. It is a fact that the seepage occurring within a soil mass causes seepage
forces, which have much greater effect than is commonly realized.
Erosion on the surface of a slope may be the cause of the removal of a certain weight of
soil, and may thus lead to an increased stability as far as mass movement is concerned. On the
other hand, erosion in the form of undercutting at the toe may increase the height of the slope,
or decrease the length of the incipient failure surface, thus decreasing the stability.
When there is a lowering of the ground water or of a freewater surface adjacent to the slope,
for example in a sudden drawdown of the water surface in a reservoir there is a decrease in the
buoyancy of the soil which is in effect an increase in the weight. This increase in weight causes
increase in the shearing stresses that may or may not be in part counteracted by the increase in
Component of weight
C
Failure
surface
(a) Infinite slope (b) An earth dam
Ground
water table
Seepage
parallel to slope
(c) Seepage below a natural slope
Lowering of water
from level A to B
Earthquake
force
(d) Sudden drawdown condition (e) Failure due to earthquake
Figure 10.1 Forces that act on earth slopes

Stability of Slopes 367
shearing strength, depending upon whether or not the soil is able to undergo compression which the
load increase tends to cause. If a large mass of soil is saturated and is of low permeability,
practically no volume changes will be able to occur except at a slow rate, and in spite of the increase
of load the strength increase may be inappreciable.
Shear at constant volume may be accompanied by a decrease in the intergranular pressure
and an increase in the neutral pressure. A failure may be caused by such a condition in which the
entire soil mass passes into a state of liquefaction and flows like a liquid. A condition of this type
may be developed if the mass of soil is subject to vibration, for example, due to earthquake forces.
The various forces that act on slopes are illustrated in Fig. 10.1.
10.2 GENERAL CONSIDERATIONS AND ASSUMPTIONS IN THE
ANALYSIS
There are three distinct parts to an analysis of the stability of a slope. They are:
1. Testing of samples to determine the cohesion and angle of internal friction
If the analysis is for a natural slope, it is essential that the sample be undisturbed.In such important
respects as rate of shear application and state of initial consolidation, the condition of testing must
represent as closely as possible the most unfavorable conditions ever likely to occur in the actual
slope.
2. The study of items which are known to enter but which cannot be accounted
for in the computations
The most important of such items is progressive cracking which will start at the top of the slope
where the soil is in tension, and aided by water pressure, may progress to considerable depth. In
addition, there are the effects of the non-homogeneous nature of the typical soil and other
variations from the ideal conditions which must be assumed.
3. Computation
If a slope is to fail along a surface, all the shearing strength must be overcome along that surface
which then becomes a surface of rupture.Any one such as ABC in Fig. 10.1(b) represents one of an
infinite number of possible traces on which failure might occur.
It is assumed that the problem is two dimensional, which theoretically requires a long length
of slope normal to the section. However, if the cross section investigated holds for a runninglength
of roughly two or more times the trace of the rupture, it is probable that the two dimensional case
holds within the required accuracy.
The shear strength of soil is assumed to follow Coulomb's law
s =c' + d tan 0"
where,
c' - effective unit cohesion
d =effective normal stress on the surface of rupture = (cr - u)
o- total normal stress on thesurface ofrupture
u - pore water pressure on the surface ofrupture
0' = effective angle of internal friction.
The item of great importance is the loss of shearing strength which many clays show when
subjected to a large shearing strain. The stress-strain curves for such clays show the stress rising
with increasing strain to a maximum value, after which it decreases and approaches an ultimate

368 Chapter 10
value which may be much less than the maximum. Since a rupture surface tends to develop
progressively rather than with all the points at the same state of strain, it is generally the ultimate
value that should be used for the shearing strength rather than the maximum value.
10.3 FACTOR OF SAFETY
In stability analysis, two types of factors of safety are normally used. They are
1. Factor of safety with respect to shearing strength.
2. Factor of safety with respect to cohesion. This is termed the factor of safety with respect to
height.
Let,
FS = factor of safety with respect to strength
F, = factor of safety with respect to cohesion
FH = factor of safety with respect to height
F, = factor of safety with respect to friction
c' = mobilized cohesion
m
0' = mobilized angle of friction
T = average value of mobilized shearing strength
s = maximum shearing strength.
The factor of safety with respect to shearing strength, F5, may be written as
s c' + <j' tan <j)'
F
>=7 = ;-
The shearing strength mobilized at each point on a failure surface may be written as
c' .
T — __ L /T
i - LJ
S
F
,
or r=c;+<7'tan0; (10.2)
c' ., tanfi
where cm -— , tanfim= -
m
p Tm
p
Actually the shearing resistance (mobilized value of shearing strength) does not develop to a
like degree at all points on an incipient failure surface. The shearing strains vary considerably and
the shearing stress may be far from constant. However the above expression is correct on the basis
of average conditions.
If the factors of safety with respect to cohesion and friction are different, we may write the
equation of the mobilized shearing resistance as
It will be shown later on that F, depends on the height of the slope. From this it may be
concluded that the factor of safety with respect to cohesion may be designated as thefactor of safety
with respect to height. This factor is denoted by FH and it is the ratio between the critical height and

the actual height, the critical height being the maximum height at which it is possible for a slope to
be stable. We may write from Eq. (10.3)
(1Q4)
H
where F^ is arbitrarily taken equal to unity.
Example 10.1
The shearing strength parameters of a soil are
c' = 26.1 kN/m2
0' = 15°
c' = 17.8 kN/m2
Calculate the factor of safety (a) with respect to strength, (b) with respect to cohesion and (c)
with respect to friction. The average intergranular pressure tf on the failure surface is 102.5 kN/m2
.
Solution
On the basis of the given data, the average shearing strength on the failure surface is
s = 26.7 + 102.5 tan 15°
= 26.7 + 102.5 x 0.268 = 54.2 kN/m2
and the average value of mobilized shearing resistance is
T= 17.8+ 102.5 tan 12°
= 17.8 + 102.5 x 0.212 = 39.6 kN/m2
F - - .L26
39.6 17.80 tan 0.212
The above example shows the factor of safety with respect to shear strength, Fs is 1.37,
whereas the factors of safety with respect to cohesion and friction are different. Consider two
extreme cases:
1. When the factor of safety with respect to cohesion is unity.
2. When the factor of safety with respect to friction is unity.
Casel
=26.70+ 102.50x0.268
9
12.90
Case 2
= 2.13
T= 39.60 = —— +102.50 tan 15C
F

370 Chapter 10
26.70
F
• + 27.50
c
12.10
We can have any combination of Fc and F, between these two extremes cited above to give
the same mobilized shearing resistance of 39.6 kN/m2
. Some of the combinations of Fc and F0 are
given below.
Combination of Fc and F^
Fc 1.00 1.26 1.37 1.50 2.20
F0 2.12 1.50 1.37 1.26 1.00
Under Case 2, the value of Fc = 2.20 when F0 - 1.0. The factor of safety FC = 2.20 is defined
as the,factor of safety with respect to cohesion.
Example 10.2
What will be the factors of safety with respect to average shearing strength, cohesion and internal
friction of a soil, for which the shear strength parameters obtained from the laboratory tests are
c' = 32 kN/m2
and 0' = 18°; the expected parameters of mobilized shearing resistance are
c'm = 21 kN/m2
and 0' = 13° and the average effective pressure on the failure plane is 110 kN/m2
.
For the same value of mobilized shearing resistance determine the following:
1. Factor of safety with respect to height;
2. Factor of safety with respect to friction when that with respect to cohesion is unity; and
3. Factor of safety with respect to strength.
Solution
The available shear strength of the soil is
s = 32 + 110 tan 18° = 32 + 35.8 = 67.8 kN/m2
The mobilized shearing resistance of the soil is
T=21 + 110 tan 13° = 21 + 25.4 = 46.4 kN/m2
_ 67.8 . .,
Factor of safety with respect to average strength, rs =—— - 1-46
46.4
32
Factor of safety with respect to cohesion, FC = —- = 1.52
_ _ tan 18° _ 0.3249 _
Factor of safety with respect to friction, F<t> - ~ TT ~~ ~ TT ~ n 2309
Factor of safety with respect to height, FH (= Fc) will be at F0 = 1 .0
. , . 32 110tanl8° , . 32
i = 46.4 = —+ - , therefore, F = - = 3.0
Fc 1.0 46.4-35.8
Factor of safety with respect to friction at F = 1 .0 is

. , . 32 110tanl8° , . ^ 35.8
r =46.4= —+ - ,therefore, F, = - = 2.49
1.0 F0 * 46.4-32
Factor of safety with respect to strength Fs is obtained when FC = F+. We may write
32 110 tan 18°
or F =1.46
10.4 STABILITY ANALYSIS OF INFINITE SLOPES IN SAND
As an introduction to slope analysis, the problem of a slope of infinite extent is of interest. Imagine
an infinite slope, as shown in Fig. 10.2, making an angle j8 with the horizontal. The soil is
cohesionless and completely homogeneous throughout. Then the stresses acting on any vertical
plane in the soil are the same as those on any other vertical plane. The stress at any point on a plane
EF parallel to the surface at depth z will be the same as at every point on this plane.
Now consider a vertical slice of material ABCD having a unit dimension normal to the page.
The forces acting on this slice are its weight W, a vertical reaction R on the base of the slice, and two
lateral forces P{ acting on the sides. Since the slice is in equilibrium, the weight and reaction are
equal in magnitude and opposite in direction. They have a common line of action which passes
through the center of the base AB. The lateral forces must be equal and opposite and their line of
action must be parallel to the sloped surface.
The normal and shear stresses on plane AB are
a' = yzcos2
fi
where cr'n = effective normal stress,
y = effective unit weight of the sand.
If full resistance is mobilized on plane AB, the shear strength, s, of the soil per Coulomb's law
is
s = af
n tan 0'
when T= s, substituting for s and tf n, we have
Figure 10.2 Stability analysis of infinite slope in sand

372
or tan /3 = tan 0'
Chapter 10
(10.5a)
Equation (10.5a) indicates that the maximum value of (3 is limited to 0' if the slope is to be
stable. This condition holds true for cohesionless soils whether the slope is completely dry or
completely submerged under water.
The factor of safety of infinite slopes in sand may be written as
p =
tanfi (10.5b)
10.5 STABILITY ANALYSIS OF INFINITE SLOPES IN CLAY
The vertical stress <Jv acting on plane AB (Fig. 10.3) where
av = yzcosfi
is represented by OC in Fig. 10.3 in the stress diagram. The normal stress on this plane is OE and
the shearing stress is EC. The line OC makes an angle (3with the cr-axis.
The Mohr strength envelope is represented by line FA whose equation is
s = c' +cr'tan^'
According to the envelope, the shearing strength is ED where the normal stress is OE.
When /3 is greater than 0' the lines OC and ED meet. In this case the two lines meet at A. As
long as the shearing stress on a plane is less than the shearing strength on the plane, there is no
danger of failure. Figure 10.3 indicates that at all depths at which the direct stress is less than OB,
there is no possibility of failure. However at a particular depth at which the direct stress is OB, the
O E B
Figure 10.3 Stability analysis of infinite slopes in clay soils

shearing strength and shearing stress values are equal as represented by AB, failure is imminent.
This depth at which the shearing stress and shearing strength are equal is called the critical depth.
At depths greater than this critical value, Fig. 10.3indicates that the shearing stress is greater than
the shearing strength but this is not possible. Therefore it may be concluded that the slope may be
steeper than 0' as long as the depth of the slope is less than the critical depth.
Expression for the Stability of an Infinite Slope of Clay of Depth H
Equation (10.2) gives the developed shearing stress as
T = c'm+(T'tan</>'m (10.6)
Under conditions of no seepage and no pore pressure, the stress components on a plane at
depth H and parallel to the surface of the slope are
r=
<j' = yHcos2
j3
Substituting these stress expressions in the equation above and simplifying, we have
c'm =YH cos2
0(tan 0- tan0'J
c'
or N = ^- = cos2
/?(tanytf-tan^) (10.7)
yti
where H is the allowable height and the term c'Jy H is a dimensionless expression called the
stability number and is designated as A^. This dimensionless number is proportional to the required
cohesion and is inversely proportional to the allowable height. The solution is for the case when no
seepage is occurring. If in Eq. (10.7) the factor of safety with respect to friction is unity,the stability
number with respect to cohesion may be written as
8)
, c
where cm= —
The stability number in Eq. (10.8) may be written as
where Hc = critical height. From Eq. (10.9), we have
Eq. (10.10) indicates that the factor of safety with respect to cohesion, Fc, is the same as the
factor of safety with respect to height FH.
If there is seepage parallel to the ground surface throughout the entire mass, with the free
water surface coinciding with the ground surface, the components of effective stresses on planes
parallel to the surface of slopes at depth H are given as [Fig. 10.4(a)].
Normal stress
(lO.lla)

374 Chapter 10
(a)
(b)
Figure 10.4 Analysis of infinite slope (a) with seepage flow through the entire
mass, and (b) with completely submerged slope.
the shearing stress
T = ysatH sin /3 cos /3
Now substitutingEqs (10.11a) and (10.lib) into equation
(lO.llb)
and simplifying, the stability expression obtained is
-^2— =cos2
0 tan 0- - -tan</>'„
Y H Y
1
sat ' sat
(10.12)
As before, if the factor of safety with respect to friction is unity, the stability number which
represents the cohesion may be writtenas
N =•
FY H Y H
c' sat 'sat ,
C/
= cos2
,tf tan^--^-
' sat
(10.13)
If the slope is completely submerged, and if there is no seepage as in Fig. 10.4(b), then
Eq. (10.13) becomes
N = = cos2
/?(tan ft ~ tan <}>') (10.14)
where y, = submerged unit weight of the soil.

Example 10.3
Find the factor of safety of a slope of infinite extent having a slope angle = 25°. The slope is made
of cohesionless soil with 0 = 30°.
Solution
Factor of safety
tan 30° 0.5774
tan/? tan 25° 0.4663
Example 10.4
Analyze the slope of Example 10.3 if it is made of clay having c' - 30 kN/m2
, 0' = 20°, e =0.65 and
Gs = 2.7 and under the following conditions: (i) when the soil is dry, (ii) when water seeps parallel
to the surface of the slope, and (iii) when the slope is submerged.
Solution
For e = 0.65 and G = 2.7
= 27x^1= = (2.7 + 0.65)x9.81=
ld
1 +0.65 /sat
1 + 0.65
yb = 10.09 kN/m3
(i) For dry soil the stability number Ns is
c
N = ——— = cos2
/?(tan/?- tan<j>') when F,=l
' d c
= (cos 25° )2
(tan 25°- tan20°) = 0.084.
c' 30
Therefore, the critical height H = - = - = 22.25 m
16.05x0.084
(ii) For seepage parallel to the surface of the slope [Eq. (10.13)]
c' 100Q
N =—-— = cos2
25° tan25°-^---tan 20° =0.2315
s
ytHc 19.9
Hc=^= 3
° =6.51 m
c
ytNs 19.9x0.2315
(iii) For the submerged slope [Eq. (10.14)]
N =cos2
25° (tan 25° - tan 20°) = 0.084
c
ybNs 10.09x0.084

376 Chapter 10
10.6 METHODS OF STABILITY ANALYSIS OF SLOPES OF FINITE
HEIGHT
The stability of slopes of infinite extent has been discussed in previous sections. A more common
problem is the one in which the failure occurs on curved surfaces. The most widely used method of
analysis of homogeneous, isotropic, finite slopes is the Swedish method based on circular failure
surfaces. Petterson (1955) first applied the circle method to the analysis of a soil failure in
connection with the failure of a quarry wall in Goeteberg, Sweden. A Swedish National
Commission, after studying a large number of failures, published a report in 1922 showing that the
lines of failure of most such slides roughly approached the circumference of a circle. The failure
circle might pass above the toe, through the toe or below it. By investigating the strength along the
arc of a large number of such circles, it was possible to locate the circle which gave the lowest
resistance to shear. This general method has been quite widely accepted as offering an
approximately correct solution for the determination of the factor of safety of a slope of an
embankment and of its foundation. Developments in the method of analysis have been made by
Fellenius (1947), Terzaghi (1943), Gilboy (1934), Taylor (1937), Bishop (1955), and others, with
the result that a satisfactory analysis of the stability of slopes, embankments and foundations by
means of the circle method is no longer an unduly tedious procedure.
There are other methods of historic interest such as the Culmann method (1875) and the
logarithmic spiral method. The Culmann method assumes that rupture will occur along a
plane. It is of interest only as a classical solution, since actual failure surfaces are invariably
curved. This method is approximately correct for steep slopes. The logarithmic spiral method
was recommended by Rendulic (1935) with the rupture surface assuming the shape of
logarithmic spiral. Though this method makes the problem statically determinate and gives
more accurate results, the greater length of time required for computation overbalances this
accuracy.
There are several methods of stability analysis based on the circular arc surface of failure. A
few of the methods are described below
Methods of Analysis
The majority of the methods of analysis may be categorized as limit equilibrium methods. The
basic assumption of the limit equilibrium approach is that Coulomb's failure criterion is satisfied
along the assumed failure surface. A free body is taken from the slope and starting from known or
assumed values of the forces acting upon the free body, the shear resistance of the soil necessary for
equilibrium is calculated. This calculated shear resistance is then compared to the estimated or
available shear strength of the soil to give an indication of the factor of safety.
Methods that consider only the whole free body are the (a) slope failure under undrained
conditions, (b) friction-circle method (Taylor, 1937, 1948) and (c) Taylor's stability number
(1948).
Methods that divide the free body into many vertical slices and consider the equilibrium of
each slice are the Swedish circle method (Fellenius, 1927), Bishop method (1955), Bishop and
Morgenstern method (1960) and Spencer method (1967). The majority of these methods are in
chart form and cover a wide variety of conditions.
10.7 PLANE SURFACE OF FAILURE
Culmann (1875) assumed a plane surface of failure for the analysis of slopes which is mainly of
interest because it serves as a test of the validity of the assumption of plane failure. In some cases
this assumption is reasonable and in others it is questionable.

Force triangle
Figure 10.5 Stability of slopes by Culmann method
The method as indicated above assumes that the critical surface of failure is a plane surface
passing through the toe of the dam as shown in Fig. 10.5.
The forces that act on the mass above trial failure planeAC inclined at angle 6 with the horizontal are
shown in the figure. The expression for the weight, W, and the total cohesion C are respectively,
W =-yLH cosec /? sin(jtf- 0)
The use of the law of sines in the force triangle, shown in the figure, gives
C _ sm(6>-f)
W ~ cos^'
Substituting herein for C and W,and rearranging we have
1
in which the subscript Q indicates that the stability number is for the trial plane at inclination 6.
The most dangerous plane is obtained by setting the first derivative of the above equation
with respect to Q equal to zero. This operation gives
where &'c is the critical angle for limiting equilibrium and the stability number for limiting
equilibrium may be written as
yHc 4sin/?cos 0'
where H is the critical height of the slope.
(10.15)

378 Chapter 10
If we write
F -— F tan
^'
c
~V' < > ~ t a n ^
where Fc and F^ are safety factors with respect to cohesion and friction respectively, Eq. (10.15)
may be modified for chosen values of c and 0' as
^=
4 sin/3 cos (/)'m (10.16)
The critical angle for any assumed values of c'm and 0'm is
1
From Eq. (10.16), the allowable height of a slope is
Example 10.5
Determine by Culmann's method the critical height of an embankment having a slope angle of 40°
and the constructed soil having c' = 630 psf, 0' = 20° and effective unit weight =114 lb/ft3
. Find the
allowable height of the embankment if F, = F, = 1 .25.
Solution
4c'sin/?cos0' 4 x 630 x sin40° cos20°
H, = ---— = - = 221 ft
y[l-cos(0-4>')] 114(l-cos20°)
For Fc =F. =1.25, c'= — =— = 504 lb/ft2
<(> m
' tan 20°
and tan #, = — - = —— =0.291, fa = 16.23°
,, , • , 4x504 sin40° cos 16.23° ^0 r
Allowable height, H = - = 128.7ft.
_ 114[l-cos(40- 16.23°)]
10.8 CIRCULAR SURFACES OF FAILURE
The investigations carried out in Sweden at the beginning of this century have clearly confirmed
that the surfaces of failure of earth slopes resemble the shape of a circular arc. When soil slips along
a circular surface, such a slide may be termed as a rotational slide. It involves downward and
outward movement of a slice of earth as shown in Fig. 10.6(a) and sliding occurs along the entire
surface of contact between the slice and its base. The types of failure that normally occur may be
classified as
1. Slope failure

2. Toe failure
3. Base failure
In slope failure, the arc of the rupture surface meets the slope above the toe. This can happen
when the slope angle /3is quite high and the soil close to the toe possesses high strength. Toefailure
occurs when the soil mass of the dam above the base and below the base is homogeneous. The base
failure occurs particularly when the base angle j3 is low and the soil below the base is softer and
more plastic than the soil above the base. The various modes of failure are shown in Fig. 10.6.
Rotational
slide
(a) Rotational slide
(b) Slope failure
(c) Toe failure
(d) Base failure
Figure 10.6 Types of failure of earth dams

380 Chapter 10
10.9 FAILURE UNDER UNDRAINED CONDITIONS (0M = 0)
A fully saturated clay slope may fail under undrained conditions (0u = 0) immediately after
construction. The stability analysis is based on the assumption that the soil is homogeneous and the
potential failure surface is a circular arc. Two types of failures considered are
1. Slope failure
2. Base failure
The undrained shear strength cu of soil is assumed to be constant with depth. A trial failure
circular surface AB with center at 0 and radius R is shown in Fig. 10.7(a) for a toe failure. The slope
AC and the chord AB make angles /3and a with the horizontal respectively. W is the weight per unit
Firm base
(a) Toe failure (b) Base failure
Figure 10.7 Critical circle positions for (a) slope failure (after Fellenius, 1927), (b)
base failure
1>
50C
40C
20°
10
90C
70°
Values of
(a)
60C
50° 50 40° 30° 20°
Values o f ?
10° 0°
Figure 10.8 (a) Relation between slope angle /3 and parameters a and Q for
location of critical toe circle when /3 is greater than 53°; (b) relation between slope
angle /3 and depth factor nd for various values of parameter nx
(after Fellenius, 1927)

length of the soil lying above the trial surface acting through the center of gravity of the mass. lo is
the lever arm, La is the length of the arc, Lc the length of the chord AB and cm the mobilized
cohesion for any assumed surface of failure.
We may express the factor of safety F^ as
(10.19)
For equilibrium of the soil mass lying above the assumed failure surface, we may write
resisting moment Mr = actuating moment Ma
The resisting moment Mf = LacmR
Actuating moment, Ma = Wlo
Equation for the mobilized c is
W10
(10.20)
Now the factor of safety F for the assumed trial arc of failure may be determined from
Eq. (10.19). This is for one trial arc. The procedure has to be repeated for several trial arcs and the
one that gives the least value is the critical circle.
If failure occurs along a toe circle, the center of the critical circle can be located by layingoff
the angles a and 26 as shown in Fig. 10.7(a). Values of a and 6 for different slope angles /3 can be
obtained from Fig. 10.8(a).
If there is a base failure as shown in Fig. 10.7(b), the trial circle will be tangential to the firm
base and as such the center of the critical circle lies on the vertical line passing through midpointM
on slope AC. The following equations may be written with reference to Fig. 10.7(b).
D x
Depth factor, nd =—, Distance factor, nx =— (10.21)
H H
Values of nx can be estimated for different values of nd and j8 by means of the chart
Fig. 10.8(b).
Example 10.6
Calculate the factor of safety against shear failure along the slip circle shown in Fig. Ex. 10.6
Assume cohesion = 40 kN/m2
, angle of internal friction = zero and the total unit weight of the
soil = 20.0 kN/m3
.
Solution
Draw the given slope ABCD as shown in Fig. Ex. 10.6. To locate the center of rotation, extend the
bisector of line BC to cut the vertical line drawn from C at point O. With O as center and OC as
radius, draw the desired slip circle.
2
Radius OC=R =36.5 m,Area BECFB = - xEFxBC
2
= - x 4 x 32.5 = 86.7 m2
Therefore W = 86.7 x 1 x 20 = 1734 kN
W acts through point G which may be taken as the middle of FE.

382 Chapter 10
s
s R = 36.5m
Figure. Ex. 10.6
From the figure we have, x = 15.2 m, and 9= 53°
3.14
Length of arc EEC =R0= 36.5 x 53° x —— =33.8 m
180
length of arc x cohesion x radius 33.8x40x36.5
Wx 1734x15.2
10.10 FRICTION-CIRCLE METHOD
Physical Concept of the Method
The principle of the method is explained with reference to the section through a dam shown in
Fig. 10.9. A trial circle with center of rotation O is shown in the figure. With center O and radius
Friction circle
Trial circular
failure surface
Figure 10.9 Principle of friction circle method

sin 0", where R is the radius of the trial circle, a circle is drawn. Any line tangent to the inner circle
must intersect the trial circle at an angle tf with R. Therefore, any vector representing an
intergranular pressure at obliquity 0' to an element of the rupture arc must be tangent to the inner
circle. This inner circle is called thefriction circle or ^-circle. The friction circle method of slope
analysis is a convenient approach for both graphical and mathematical solutions. It is given this
name because the characteristic assumption of the method refers to the 0-circle.
The forces considered in the analysis are
1. The total weight W of the mass above the trial circle acting through the center of mass. The
center of mass may be determined by any one of the known methods.
2. The resultant boundary neutral force U. The vector U may be determined by a graphical
method from flownet construction.
3. The resultant intergranularforce, P, acting on the boundary.
4. The resultant cohesive force C.
Actuating Forces
The actuating forces may be considered to be the total weight W and the resultant boundary force U
as shown in Fig. 10.10.
The boundary neutral force always passes through the center of rotation O.The resultant ofW
and U, designated as Q, is shown in the figure.
Resultant Cohesive Force
Let the length of arc AB be designated as La, the length of chord AB by Lc. Let the arc length La be
divided into a number of small elements and let the mobilized cohesive force on these elements be
designated as Cr C2, C3, etc. as shown in Fig. 10.11. The resultant of all these forces is shown by
the force polygon in the figure. The resultant is A'B' which is parallel and equal to the chord length
AB. The resultant of all the mobilized cohesional forces along the arc is therefore
C = c'L
Figure 10.10 Actuating forces

384 Chapter 10
(a) Cohesive forces on a trial arc (b) Polygon of forces
Figure 10.11 Resistant cohesive forces
We maywrite c'm - —
c
wherein c'= unit cohesion, FC = factor of safety with respect to cohesion.
The line of action of C may be determined by moment consideration. The moment of the total
cohesion is expressed as
c' L R = c' L I
m a m c a
where l = moment arm. Therefore,
(10.22)
It is seen that the line of action of vector C is independent of the magnitude of c'm.
Resultant of Boundary Intergranular Forces
The trial arc of the circle is divided into a number of small elements. Let Pv P2, Py etc. be the
intergranular forces acting on these elements as shown in Fig. 10.12. The friction circle is drawn
with a radius of R sin (j/m
where
The lines of action of the intergranular forces Pr P2, Py etc. are tangential to the friction
circle and make an angle of 0'm at the boundary. However, the vector sum of any two small
forces has a line of action through point D, missing tangency to the 0'm-circle by a small
amount. The resultant of all granular forces must therefore miss tangency to the 0'm-circle by
an amount which is not considerable. Let the distance of the resultant of the granular force P
from the center of the circle be designated as KR sin 0' (as shown in Fig. 10.12). The

KRsin<p'n
Figure 10.12 Resultant of intergranular forces
magnitude of K depends upon the type of intergranular pressure distribution along the arc. The
most probable form of distribution is the sinusoidal distribution.
The variation of K with respect to the central angle a'is shown in Fig. 10.13. The figure also
gives relationships between of and K for a uniform stress distribution of effective normal stress
along the arc of failure.
The graphical solution based on the concepts explained above is simple in principle. For the
three forces Q, C and P of Fig. 10.14 to be in equilibrium, P must pass through the intersection of
1.20
1.16
1.12
1.08
1.04
1.00
ox
£
^
71
str
^
Cent
For i
essc
/
ral angle
anifo
istrih
/
^
rm
>utior
j
/
s<
i —S
/
/
/
s
/
y
/
For
tress
/
'
/
sinus
distr
/
/
/
oida
ibuti(
J
/
^n
20 40 60 80
Central angle a' in degrees
100 120
Figure 10.13 Relationship between K and central angle a'

386 Chapter 10
Figure 10.14 Force triangle for the friction-circle method
the known lines of action of vectors Q and C. The line of action of vector P must also be tangent to
the circle of radius KR sin 0' . The value of K may be estimated by the use of curves given in
Fig. 10.13, and the line of action offeree P may be drawn as shown in Fig. 10.14. Since the lines of
action of all three forces and the magnitude of force Q are known, the magnitude of P and C may-be
obtained by the force parallelogram construction that is indicated in the figure. The circle of radius
of KR sin 0' is called the modified friction circle.
T
rn j j
Determination of Factor of Safety With Respect to Strength
Figure 10.15(a) is a section of a dam. AB is the trial failure arc. The force Q, the resultant of W
and U is drawn as explained earlier. The line of action of C is also drawn. Let the forces Q and C
D
(a) Friction circle (b) Factor of safety
Figure 10.15 Graphical method of determining factor of safety with respect to
strength

meet at point D. An arbitrary first trial using any reasonable $m value, which will be designated
by 0'ml is given by the use of circle 1 or radius KR sin <j)'ml. Subscript 1 is used for all other
quantities of the first trial. The force Pl is then drawn through D tangent to circle 1. Cl is parallel
to chord and point 1 is the intersection of forces C{ and Pr The mobilized cohesion is equal
c'm]Lc. From this the mobilized cohesion c'ml is evaluated. The factors of safety with respect to
cohesion and friction are determined from the expressions
c' tanfl'
F' =——, and F*,
These factors are the values used to plot point 1 in the graph in Fig. 10.15(b). Similarly
other friction circles with radii KR sin <j/m2, KR sin 0'm3. etc. may be drawn and the procedure
repeated. Points 2, 3 etc. are obtained as shown in Fig.10.15(b). The 45° line, representing
Fc = F., intersects the curve to give the factor of safety Fs for this trial circle.
Several trial circles must be investigated in order to locate the critical circle, which is the one
having the minimum value of F5.
Example 10.7
An embankment has a slope of 2 (horizontal) to 1 (vertical) with a height of 10 m. It is made of a
soil having a cohesion of 30 kN/m2
, an angle of internal friction of 5° and a unit weight of
20 kN/m3
. Consider any slip circle passing through the toe. Use the friction circle method to find
the factor of safety with respect to cohesion.
Solution
Refer to Fig. Ex. 10.7. Let EFB be the slope and AKB be the slip circle drawn with center O and
radius R =20 m.
Length of chord AB =Lc = 27 m
Take J as the midpoint of AB, then
Area AKBFEA = area AKBJA + area ABEA
= -ABxJK +-ABxEL
3 2
= -x27 x5.3+ - x27 x2.0 = 122.4 m2
3 2
Therefore the weight of the soil mass = 122.4 x 1 x 20 = 2448 kN
It will act through point G, the centroid of the mass which can be taken as the mid point of
FK.
Now, 0=85°,
314
Length of arc AKB = L = RO = 20 x 85x— = 29.7 m
6
180
L 29.7
Moment arm of cohesion, / = R— = 20 x —— = 22 m
L
c 21
From center O, at a distance /fl, draw the cohesive force vector C, which is parallel to the
chord AB. Now from the point of intersection of C and W, draw a line tangent to the friction circle

388 Chapter 10
1.74m
//=10m
Figure Ex. 10.7
drawn at 0 with a radius of R sin 0' = 20 sin 5° = 1 .74 m. This line is the line of action of the third
force F.
Draw a triangle of forces in which the magnitude and the direction for W is known and only
the directions of the other two forces C and F are known.
Length ad gives the cohesive force C = 520 kN
Mobilized cohesion,
c' = - = — = 17.51 kN/m2
m
L 29.7
Therefore the factor of safety with respect to cohesion, Fc, is
F =11= ^=1.713
FC will be 1 .7 13 if the factor of safety with respect to friction, F^ - 1 .0
tan5c
If, F = 1.5, then 0' =
F.
= 0.058 rad; or 0' = 3.34°

The new radius of the friction circle is
r{ =R sin 0'm = 20 x sin 3.3° = 1.16 m.
The direction of F changes and the modified triangle of force abd' gives,
cohesive force = C = length ad' = 600 kN
C 600
Mobilised cohesino, c'm =~— - - 20.2 kN/mr
LJ Z*yI /
c' 30
Therefore, F = — = = 1.5
c
c' 20.2
10.1 1 TAYLOR'S STABILITYNUMBER
If the slope angle j8, height of embankment H, the effective unit weight of material y, angle of
internal friction </>', and unit cohesion c' are known, the factor of safety may be determined. In order
to make unnecessary the more or less tedious stability determinations, Taylor (1937) conceived the
idea of analyzing the stability of a large number of slopes through a wide range of slope angles and
angles of internal friction, and then representing the results by an abstract number which he called
the "stability number". This number is designated as A^. The expression used is
From this the factor of safety with respect to cohesion may be expressed as
F
- = 7 <10
-24
>
Taylor published his results in the form of curves which give the relationship between Ns and
the slope angles /? for various values of 0' as shown in Fig. 10.16. These curves are for circles
passing through the toe, although for values of 13 less than 53°, it has been found that the most
dangerous circle passes below the toe. However, these curves may be used without serious error for
slopes down to fi = 14°. The stability numbers are obtained for factors of safety with respect to
cohesion by keeping the factor of safety with respect to friction (FJ equal to unity.
In slopes encountered in practical problems, the depth to which the rupture circle may extend
is usually limited by ledge or other underlying strong material as shown in Fig. 10.17. The stability
number Ns for the case when 0" = 0 is greatly dependent on the position of the ledge. The depth at
which the ledge or strong material occurs may be expressed in terms of a depth factor nd which is
defined as
»rf=;| (10-25)
where D - depth of ledge below the top of the embankment, H =height of slope above the toe.
For various values of nd and for the 0 = 0 case the chart in Fig. 10.17 gives the stability
number NS for various values of slope angle ft. In this case the rupture circle may pass through the
toe or below the toe. The distance jc of the rupture circle from the toe at the toe level may be
expressed by a distance factor n which is defined as

•a Stability number, Ns
CD
H °
|_cu
o a
cr
-* CD
<° o
^ ^
C
CD
_j
O
O)
0)
o"
V)*
Q)
CT
(Q
C
Stability number, N,.
Q)
-. <.
Q) O
—f> -i
r-t- >
CD cn
~" w
—J r-+
Q) 0)
<
. E
o ^~
CD C
co 3
->J cr
'** CD
-^
cn
oee

The chart in Fig. 10.17 shows the relationship between nd and nx. If there is a ledge or other
stronger material at the elevation of the toe, the depth factor nd for this case is unity.
Factor of Safety with Respect to Strength
The development of the stability number is based on the assumption that the factor of safety with
respect to friction F,, is unity. The curves give directly the factor of safety Fc with respect to
cohesion only. If a true factor of safety Fs with respect to strength is required, this factor should
apply equally to both cohesion and friction. The mobilized shear strength may therefore be
expressed as
s c' a' tan (/)'
In the above expression, we may write
— =c'm, tan (f>'m =—=— , or #, = — (approx.) (10.27)
S 5 S
c'm and tf m may be described as average values of mobilized cohesion and friction respectively.
Example 10.8
The following particulars are given for an earth dam of height 39 ft. The slope is submerged and the
slope angle j3 = 45°.
Yb = 69 lb/ft3
c' = 550 lb/ft2
0' = 20°
Determine the factor of safety FS.
Solution
Assume as a first trial Fs = 2.0
20
<t>'m
=
Y= 10
° (approx.)
For (j)'m = 10°, and (3 = 45° the value of Ns from Fig. 10.16 is 0.11, we may write
c'
From Eq. (10.23) N = - , substituting
55Q
2x69x#
or H = — =36.23ft
2x69x0.11
20
If F = 1.9, $ =— =10.53° and N = 0.105
5
19 *

392 Chapter 10
.40ft
1.9x69x0.105
The computed height 40 ft is almost equal to the given height 39 ft. The computed factor of
safety is therefore 1 .9.
Example 10.9
An excavation is to be made in a soil deposit with a slope of 25° to the horizontal and to a depth of
25 meters. The soil has the following properties:
c'= 35kN/m2
, 0' = 15° and 7= 20 kN/m3
1. Determine the factor of safety of the slope assuming full friction is mobilized.
2. If the factor of safety with respect to cohesion is 1.5, what would be the factor of safety
with respect to friction?
Solution
1. For 0' = 15° and (3 = 25°, Taylor's stability number chart gives stability number Ns = 0.03.
-233
0.03x20x25
2. For F = 1.5, N = --- --—- =0.047
J
FcxyxH 1.5x20x25
For A^ = 0.047 and (3 = 25°, we have from Fig. 10.16, 0'm = 13
tan0' tan 15° 0.268
Therefore, F, = -— = - = - = 1.16
0
tan0 tan 13° 0.231
Example 10.10
An embankment is to be made from a soil having c' = 420 lb/ft2
, 0' = 18° and y= 121 lb/ft3
. The
desired factor of safety with respect to cohesion as well as that with respect to friction is 1.5.
Determine
1. The safe height if the desired slope is 2 horizontal to 1 vertical.
2. The safe slope angle if the desired height is 50 ft.
Solution
, 0.325
tan 0' =tan18°=0.325, 0'm - tan ' —- =12.23°
1. For 0' = 12.23° and (3= 26.6° (i.e., 2 horizontal and 1 vertical) the chart gives Ns = 0.055
c' 420
Therefore, 0.055 =
FcyH 1.5 x 121 xH

Therefore, # . =
420
2. Now, NS = •
safe
1.5x121x0.055
420
= 42 ft
= 0.046
FcyH 1.5x121x50
For N = 0.046 and 0' = 12.23°, slope angle P = 23.5C
10.12 TENSION CRACKS
If a dam is built of cohesive soil, tension cracks are usually present at the crest. The depth of such
cracks may be computed from the equation
r (10.28)
where z0 = depth of crack, c' = unit cohesion, y = unit weight of soil.
The effective length of any trial arc of failure is the difference between the total length of arc
minus the depth of crack as shown in Fig. 10.18.
10.13 STABILITY ANALYSIS BY METHOD OF SLICES FOR
STEADY SEEPAGE
The stability analysis with steady seepage involves the development of the pore pressure head
diagram along the chosen trial circle of failure. The simplest of the methods for knowing the pore
pressure head at any point on the trial circle is by the use of flownets which is described below.
Determination of Pore Pressure with Seepage
Figure 10.19 shows the section of a homogeneous dam with an arbitrarily chosen trial arc.There is
steady seepage flow through the dam as represented by flow and equipotential lines. From the
equipotential lines the pore pressure may be obtained at any point on the section. For example at
point a in Fig. 10.19 the pressure head is h. Point c is determined by setting the radial distance ac
Tension crack
Effective length of
trial arc of failure
Figure 10.18 Tension crack in dams built of cohesive soils

394 Chapter 10
Trial circle
- ' 'R= radius /
of trial circle/'
d/s side /
Phreatic line
Piezometer
Pressure head
at point a - h
Discharge face
- Equipotential line x
r ---- -'
Pore pressure head diagram -/
Figure 10.19 Determination of pore pressure with steady seepage
equal to h. A number of points obtained in the same manner as c give the curved line through c
which is a pore pressure head diagram.
Method of Analysis (graphical method)
Figure 10.20(a) shows the section of a dam with an arbitrarily chosen trial arc. The center of
rotation of the arc is 0. The pore pressure acting on the base of the arc as obtained from flow nets is
shown in Fig. 10.20(b).
When the soil forming the slope has to be analyzed under a condition where full or partial
drainage takes place the analysis must take into account both cohesive and frictional soil properties
based on effective stresses. Since the effective stress acting across each elemental length of the
assumed circular arc failure surface must be computed in this case, the method of slices is one of
the convenient methods for this purpose. The method of analysis is as follows.
The soil mass above the assumed slip circle is divided into a number of vertical slices of equal
width. The number of slices may be limited to a maximum of eight to ten to facilitate computation.
The forces used in the analysis acting on the slices are shown in Figs. 10.20(a) and (c). The forces
are:
1. The weight W of the slice.
2. The normal and tangential components of the weight W acting on the base of the slice.
They are designated respectively as N and T.
3. The pore water pressure U acting on the base of the slice.
4. The effective frictional and cohesive resistances acting on the base of the slice which is
designated as S.
The forces acting on the sides of the slices are statically indeterminate as they depend on the
stress deformation properties of the material, and we can make only gross assumptions about their
relative magnitudes.
In the conventionalslice method of analysisthe lateral forces are assumed equal on both sides
of the slice. This assumption is not strictly correct. The error due to this assumption on the mass as
a whole is about 15 percent (Bishop, 1955).

(a) Total normal and tangential components
B ~ - - ^ C
(b) Pore-pressure diagram
Trial failure
surface
fl
/ 7" U} = «,/,
Pore-pressure
diagram
U2 = M2/2
U3 = M3/3
(c) Resisting forces on the base of slice (d) Graphical representation of all the forces
Figure 10.20 Stability analysis of slope by the method of slices

396 Chapter 10
The forces that are actually considered in the analysis are shown in Fig. 10.20(c). The various
components may be determined as follows:
1. The weight, W, of a slice per unit length of dam may be computed from
W=yhb
where, y = total unit weight of soil, h =average height of slice, b - width of slice.
If the widths of all slices are equal, and if the whole mass is homogeneous, the weight W
can be plotted as a vector AB passing through the center of a slice as in Fig. 10.20(a). AB
may be made equal to the height of the slice.
2. By constructing triangle ABC, the weight can be resolved into a normal component N and
a tangential component T. Similar triangles can be constructed for all slices. The tangential
components of the weights cause the mass to slide downward. The sum of all the weights
cause the mass_ to slide downward. The sum of all the tangential components may be
expressed as T= I.T. If the trial surface is curved upward near its lower end, the tangential
component of the weight of the slice will act in the opposite direction along the curve. The
algebraic sum of T should be considered.
3. The average pore pressure u acting on the base of any slice of length / may be found from
the pore pressure diagram shown in Fig. 10.20(b). The total pore pressure, U, on the base of
any slice is
U=ul
4. The effective normal pressure N' acting on the base of any slice is
N'=N- t/[Fig. 10.20(c)]
5. The frictional force Ff
acting on the base of any slice resisting the tendency of the slice to
move downward is
F =(N - U) tan 0'
where 0' is the effective angle of friction. Similarly the cohesive force C" opposing the
movement of the slice and acting at the base of the slice is
where c is the effective unit cohesion. The total resisting force S acting on the base of the
slice is
S = C +F' =c'l +(N - U) tan 0'
Figure 10.20(c) shows the resisting forces acting on the base of a slice.
The sum of all the resisting forces acting on the base of each slice may be expressed as
Ss =c'I,l +tan 0' I(W- £/) = c'L +tan 0' X(N - U)
where £/ = L = length of the curved surface.
The moments of the actuating and resisting forces about the point of rotation may be
written as follows:
Actuating moment = R~LT
Resisting moment = R[c'L + tan 0' £(jV - U)]
The factor of safety F? may now be written as
(10.29)

The various components shown in Eq. (10.29) can easily be represented graphically as
shown in Fig. 10.20(d). The line AB represents to a suitable scale Z,(N - U). BC is drawn
normal to AB at B and equal to c'L + tan 0' Z(N - U).The line AD drawn at an angle 0'to
AB gives the intercept BD on BC equal to tan 0'Z(N- U).The length BE on BC is equal to
IT.Now
F =
BC
BE
(10.30)
Centers for Trial Circles Through Toe
The factor of safety Fs as computed and represented by Eq. (10.29) applies to one trial circle. This
procedure is followed for a numberof trial circles until one finds the one for which the factor of safety
is the lowest. This circle that gives the least Fs is the one most likely to fail. The procedure is quite
laborious. The number of trial circles may be minimized if one follows the followingmethod.
For any given slope angle /3 (Fig. 10.21), the center of the first trial circle center O may be
determined as proposed by Fellenius (1927). The direction angles aA and aB may be taken from
Table 10.1. For the centers of additional trial circles, the procedure is as follows:
Mark point C whose position is as shown in Fig. 10.21. Join CO. The centers of additional
circles lie on the line CO extended. This method is applicable for a homogeneous (c - </>) soil. When
the soil is purely cohesive and homogeneous the direction angles given in Table 10.1 directly give
the center for the critical circle.
Centers for Trial Circles Below Toe
Theoretically if the materials of the dam and foundation are entirely homogeneous, any practicable
earth dam slope may have its critical failure surface below the toe of the slope. Fellenius found that
the angle intersected at 0 in Fig. 10.22 for this case is about 133.5°. Tofind the center for the critical
circle below the toe, the following procedure is suggested.
Locus of centers
of critical circles
Curve of factor
of safety
Figure 10.21 Location of centers of critical circle passing through toe of dam

398 Chapter 10
Figure 10.22 Centers of trial circles for base failure
Table 10.1 Direction angles a°A and a°ofor centers of critical circles
Slope Slope angle Direction angles
0.6:1
1 :1
1.5: 1
2: 1
3: 1
5 : 1
60
45
33.8
26.6
18.3
11.3
29
28
26
25
25
25
40
37
35
35
35
37
Erect a vertical at the midpoint M of the slope. On this vertical will be the center O of the first
trial circle. In locating the trial circle use an angle (133.5°) between the two radii at which the circle
intersects the surface of the embankment and the foundation. After the first trial circle has been
analyzed the center is some what moved to the left, the radius shortened and a new trial circle drawn
and analyzed. Additional centers for the circles are spotted and analyzed.
Example 10.11
An embankment is to be made of a sandy clay having a cohesion of 30 kN/m2
, angle of internal
friction of 20° and a unit weight of 18 kN/m3
. The slope and height of the embankment are 1.6 : 1
and 10m respectively. Determine the factor of safety by using the trial circle given in Fig. Ex. 10.11
by the method of slices.
Solution
Consider the embankment as shown in Fig. Ex.10.11. The center of the trial circle O is selected by
taking aA = 26° and aB =35° from Table 10.1. The soil mass above the slip circle is divided into 13
slices of 2 m width each. The weight of each slice per unit length of embankment is given by W=
haby;, where ha = average height of the slice, b = width of the slice, yt = unit weight of the soil.
The weight of each slice may be represented by a vector of height ha if b and y, remain the
same for the whole embankment.The vectors values were obtained graphically. The height vectors

Figure Ex. 10.11
may be resolved into normal components hn and tangential components h{. The values of ha, hn and
ht for the various slices are given below in a tabular form.
Values of hal hn and /?,
o / v r
Slice No.
1
2
3
4
5
6
7
ha(m)
1.8
5.5
7.8
9.5
10.6
11.0
10.2
hn(m)
0.80
3.21
5.75
7.82
9.62
10.43
10.20
ht(m]
1.72
4.50
5.30
5.50
4.82
3.72
2.31
Slice No.
8
9
10
11
12
13
ha(m)
9.3
8.2
6.8
5.2
3.3
1.1
hn(m)
9.25
8.20
6.82
5.26
3.21
1.0
ht(m)
1.00
-0.20
-0.80
-1.30
-1.20
-0.50
The sum of these components hn and ht may be converted into forces ZN and Irrespectively
by multiplying them as given below
Sfcn = 81.57m, Uit = 24.87m
Therefore, ZN = 81.57x 2 x 18 = 2937 kN
Zr = 24.87 x2x 18= 895kN
Length of arc = L = 31.8 m
'L + tonfiZN 30x31.8 + 0.364x2937
Factor of safety =
895
=2.26

400
10.14 BISHOP'S SIMPLIFIED METHOD OF SLICES
Chapter 10
Bishop's method of slices (1955) is useful if a slope consists of several types of soil with different
values of c and 0 and if the pore pressures u in the slope are known or can be estimated. The method
of analysis is as follows:
Figure 10.23 gives a section of an earth dam having a sloping surface AB. ADC is an assumed
trial circular failure surface with its center at O. The soil mass above the failure surface is divided
into a number of slices. The forces acting on each slice are evaluated from limit equilibrium of the
slices. The equilibrium of the entire mass is determined by summation of the forces on each of the
slices.
Consider for analysis a single slice abed (Fig. 10.23a) which is drawn to a larger scale in
Fig. 10.23(b). The forces acting on this slice are
W = weight of the slice
N = total normal force on the failure surface cd
U = pore water pressure = ul on the failure surface cd
FR = shear resistance acting on the base of the slice
Er E2 - normal forces on the vertical faces be andad
Tr T2 = shear forces on the vertical faces be and ad
6 = the inclination of the failure surface cd to the horizontal
The system is statically indeterminate. An approximate solution may be obtained by
assuming that the resultant of £, and T^ is equal to that of E2 and T2, and their lines of action
coincide. For equilibrium of the system, the following equations hold true.
O
(a) (b)
Figure 10.23 Bishop's simplified method of analysis

N=Wcos6
(10.31)
where F( = tangential component of W
The unit stresses on the failure surface of length, /, may be expressed as
Wcos6
normal stress, <rn = -
Wsin0 (10.32)
shear stress, rn = -
The equation for shear strength, s, is
s =c' + cr'tan^' = c' + (cr-u)tan0'
where rf =effective normal stress
c' - effective cohesion
(ft = effective angle of friction
u = unit pore pressure
The shearing resistance to sliding on the base of the slice is
si =c'l +(Wcos 9- ul) tan$
where ul = U, the total pore pressure on the base of the slice (Fig 10.23b)
d = F
At - rR
The total resisting force and the actuating force on the failure surface ADC may be expressed
as
Total resisting force FR is
FR= [c7 + (Wcos0-M/)tan0'] (10.33)
Total actuating force Ft is
Ft = Wsm0 (10.34)
The factor of safety Fs is then given as
F
Eq. (10.35) is the same as Eq. (10.29) obtained by the conventional method of analysis.
Bishop (1955) suggests that the accuracy of the analysis can be improved by taking into
account the forces E and Ton the vertical faces of each slice. For the element in Fig. 10.23(b), we
may write an expression for all the forces acting in the vertical direction for the equilibrium
condition as
N' co&0 = W +(T^ -T2)-ulcos0- FR sin# (10.36)
If the slope is not on the verge of failure (Fs > 1), the tangential force Ft is equal to the
shearing resistance FR on cd divided by Fg.

402 Chapter 10
c'l
(10.37)
where, N'=N-U,andU= ul.
Substituting Eq. (10.37) into Eq. (10.36) and solving for N we obtain
c'l
— sin<9
F
cos <9 +
tan 0' sin6
F..
(10.38)
where, AT= T{ - Tr
For equilibrium of the mass above the failure surface, we have by taking moments about O
Wsin0R = FRR (10.39)
By substitutingEqs.(10.37) and (10.38) into Eq. (10.39) and solving we obtain an expression
forF as
F
where,
tan (/>' sin 9
F
(10.40)
(10.41)
The factor of safety FS is present in Eq. (10.40) on both sides. The quantity AT= T{ - T2 has
to be evaluated by means of successive approximation . Trial values of E^ and Tl that satisfy the
equilibrium of each slice, and the conditions
1.6
1.4
1.2
1.0
mf) = cos 6 + (sin6 tan d) )/F
i i i
Note: 0 is + when slope of failure
arc is in the same quadrant
as ground slope.
0.6
-40 -30 -20 -10 0 10 20
Values of 6 degrees
30 40
Figure 10.24 Values of mfi (after Janbu et al., 1956)

(El-E2) =Q and (rl-T2) = 0
are used. The value of Fs may then be computed by first assuming an arbitrary value for Fs. The
value of Fs may then be calculated by making use of Eq. (10.40). If the calculated value of Fv differs
appreciably from the assumed value, a second trial is made and the computation is repeated.
Figure 10.24developed by Janbu et al. (1956) helps to simplify the computation procedure.
It is reported that an error of about 1percent will occur if we assume Z(Tj - T"2) tan0'= 0.But
if we use the conventional method of analysis using Eq. (10.35) the error introduced is about
15 percent (Bishop, 1955).
10.15 BISHOP AND MORGENSTERN METHOD FOR SLOPE ANALYSIS
Equation (10.40) developed based on Bishop's analysis of slopes, contains the term pore pressure
u. The Bishop and Morgenstern method (1960) proposes the following equation for the evaluation
of u
yh (10.42)
where, u = pore water pressure at any point on the assumed failure surface
Y= unit weight of the soil
h = the depth of the point in the soil mass below the ground surface
The pore pressure ratio ru is assumed to be constant throughout the cross-section, which is
called a homogeneous pore pressure distribution. Figure 10.25shows the various parameters used
in the analysis.
The factor of safety F is defined as
F_ = m - nr,. (10.43)
where, m, n = stabilitycoefficients.
The m and n values may be obtained either from charts in Figs. B.1to B.6 or Tables B1 to B6
in Appendix B. The depth factor given in the charts or tables is as per Eq. (10.25), that is nd = DIH,
where H = height of slope, and D = depth of firm stratum from the top of the slope. Bishop and
Morgenstern (1960) limited their charts (or tables) to values of c'ly H equal to 0.000, 0.025, and
0.050.
Center of failure surface
Failure surface
y = unit weight of soil
/^^^^^^^^//^f^^^
Figure 10.25 Specifications of parameters for Bishop-Morgenstern method of
analysis

404 Chapter 10
Extension of the Bishop and Morgenstern Slope Stability Charts
As stated earlier, Bishop and Morgenstern (1960) charts or tables cover values of c'lyH equal to
0.000, 0.025, and 0.050 only.These charts do not cover the values that are normally encountered in
natural slopes. O' Connor and Mitchell (1977) extended the work of Bishop and Morgenstern to
cover values of c'lyH equal to 0.075 and 0.100 for various values of depth factors nd. The method
employed is essentially the same as that adopted by the earlier authors. The extended values are
given in the form of charts and tables from Figs. B.7 to B.14 and Tables B7 to B14 respectively in
Appendix B.
Method of Determining Fs
1. Obtain the values of ru and clyH
2. From the tables in Appendix B, obtain the values of m and n for the known values ofc/yH,
0 and /3, and for nd - 0, 1, 1.25 and 1.5.
3. Using Eq. (10.43), determine Fs for each value of nd.
4. The required value of Fs is the lowest of the values obtained in step 3.
Example 10.12
Figure Ex. 10.12gives a typical section of a homogeneous earth dam.The soil parameters are:
0' = 30°, c' = 590 lb/ft2
, and y = 120 lb/ft3
. The dam has a slope 4:1 and a pore pressure ratio
ru = 0.5. Estimate the factor of safety Fs by Bishop and Morgenstern method for a height of dam
#=140 ft.
Solution
Height of dam H= 140ft
c' 590
120x140
= 0.035
Given: 0' = 30°,slope 4:1 and ru = 0.5.
Since c'lyH = 0.035, and nd = 1.43for H = 140 ft, the Fs for the dam lies between c'lyH
0.025 and 0.05 and nd between 1.0 and 1.5. The equation for Fs is
= m-nr
Using the Tables in Appendix B, the following table can be prepared for the given values of
c'lyH, 0, and /3.
0'=30°
c' = 590psf
y - 120 pcf
/•„ =0.50
D = 200 ft
Alluvium (same properties as above)
Figure Ex. 10.12

From Tables B2 and B3 for c'/yH =0.025
n
d
1.0
1.25
From Table B4, B5
n
d
1.0
1.25
1.50
m
2.873
2.953
and B6for c'ljH - 0.05
m
3.261
3.221
3.443
n
2.622
2.806
n
2.693
2.819
3.120
F,
1.562
1.55
F,
1.915
1.812
1.883
Lowest
Lowest
Hence nd = 1.25 is the more critical depth factor. The value of Fs for c'lyH = 0.035 lies
between 1.55 (for c'lyH = 0.025) and 1.812 (for c'lyH = 0.05). By proportion F = 1.655.
10.16 MORGENSTERN METHOD OF ANALYSIS FOR RAPID
DRAWDOWN CONDITION
Rapid drawdown of reservoir water level is one of the critical states in the design of earth dams.
Morgenstern (1963) developed the method of analysis for rapid drawdown conditions based on the
Bishop and Morgenstern method of slices. The purpose of this method is to compute the factor of
safety during rapid drawdown, which is reduced under no dissipation of pore water pressure. The
assumptions made in the analysis are
1. Simple slope of homogeneous material
2. The dam rests on an impermeable base
3. The slope is completely submerged initially
4. The pore pressure does not dissipate during drawdown
Morgenstern used the pore pressure parameter 5 as developed by Skempton (1954) which
states
5 = — (10.45)
where cr, = y h
j- total unit weight of soil or equal to twice the unit weight of water
h = height of soil above the lower level of water after drawdown
The charts developed take into account the drawdown ratio which is defined as
(10.46)
where Rd = drawdown ratio
// = height of drawdown
H = height of dam (Fig. 10.26)
All the potential sliding circles must be tangent to the base of the section.

406 Chapter 10
Full reservoir level
" Drawdown
/level H
Figure 10.26 Dam section for drawdown conditions
The stability charts are given in Figs. 10.27 to 10.29 covering a range of stability numbers
c'/yH from 0.0125 to 0.050. The curves developedare for the values of 0'of 20°, 30°, and 40° for
different values of B.
PL,
0.2 0.4 0.6 _0.8
Drawdown ratio H/H
(a) 0 = 2:1
1.0 0 0.2 0.4 0.6 _0.8
Drawdown ratio H/H

<P
40°
30°
20°
0.2 0.4 0.6 _0.8
Drawdown ratio H/H
1.0
Figure 10.27
0.2 0.4 0.6 _0.8 1.0
Drawdown ratio H/H
(d) ft = 5:1
Drawdown stability chart for c'/yH = 0.0125 (after Morgenstern,
1963)

40°
30°
20°
0.2 0.4 0.6 _0.8 1.0
Drawdown ratio H/H
(a) ft = 2:1
40°
30°
20°
0 0.2 0.4 0.6 _0.8 1.0
Drawdown ratio H/H
(b) ft = 3:1
0.2 0.4 0.6 _0.8
Drawdown ratio H/H
(c) ft = 4:1
<P
40°
30°
20°
1.0
i*,
>*
UH
(d) 0 = 5:
<P
40°
30°
20°
0 0.2 0.4 0.6 _0.8 1.0
Drawdown ratio H/H
Figure 10.28 Drawdown stability chart for c'lyH = 0.025 (after Morgenstern,
1963)
Example 10.13
It is required to estimate the minimum factor of safety for the complete drawdown of the section
shown in Fig. Ex. 10.13 (Morgenstern, 1963)
.*._./:
Water level before
drawdown
Water level after
drawdown
Figure Ex. 10.13

408 Chapter 10
Solution
From the data given in the Fig. Ex. 10.13
N =— =
312
= 0.025
yH. 124.8x100
From Fig. 10.28, for W = 0.025, 0= 3:1, </>' = 30°, and H/H = 1,
Fs = 1.20
It is evident than the critical circle is tangent to the base of the dam and no other level need be
investigated since this would only raise the effective value of NS resulting in a higher factor of
safety.
10.17 SPENCER METHOD OF ANALYSIS
Spencer (1967) developed his analysis based on the method of slices of Fellenius (1927) and
Bishop (1955). The analysis is in terms of effective stress and satisfies two equations of
X
0.2 0.4 0.6 _0.8
Drawdown ratio H/H
0 0.2 0.4 0.6 _0.8
Drawdown ratio H/H
40°
30°
20°
1.0 0.2 0.4 0.6 _0.8
Drawdown ratio H/H
n- 2
k

.
l

X
X

X
X
X

x
.^"•^
X
x^
»^_
—
——E^M
40°
30°
20°
0 0.2 0.4 0.6 _0.8
Drawdown ratio H/H
1.0
(c) ft = 4:1 (d) ft = 5:1
Figure 10.29 Drawdown stability chart for c'lyH = 0.05 (after Morgenstern, 1963)

equilibrium, the first with respect to forces and the second with respect to moments. The interslice
forces are assumed to be parallel as in Fig. 10.23. The factor of safety Ff is expressed as
Shear strength available
F = CIO 47)
5
Shear strength mobilized ' '
The mobilized angle of shear resistance and other factors are expressed as
(10.48)
u
pore pressure ratio, r =— nn 49)
yh ^ ' '
c'
Stability factor, NS=—— (10.50)
The charts developed by Spencer for different values of Ns, §'m and ru are given in Fig. 10.30.
The use of these charts will be explained with worked out examples.
Example 10.14
Find the slope corresponding to a factor of safety of 1.5 for an embankment 100 ft high in a soil
whose properties are as follows:
c' = 870 Ib/sq ft, y= 120 Ib/ft3, </>' = 26°, ru = 0.5
Solution (by Spencer's Method)
N =^L= 870
5
Fsytl 1.5x120x100
t ., tanf 0.488 _ „ _
tan 0 = -— = - =0.325
F 1.5
Referring to Fig. 10.30c, for which r =0.5, the slope corresponding to a stability number
of 0.048 is 3:1.
Example 10.15
What would be the change in strength on sudden drawdown for a soil element at point P which is
shown in Fig. Ex. 10.15? The equipotential line passing through this element represents loss of
water head of 1.2 m. The saturated unit weight of the fill is 21 kN/m3
.
Solution
The data given are shown in Fig. Ex. 10.15. Before drawdown,
The stresses at point P are:
% = /A+
nA = 9.81 x 3 + 21x 4 = 113 kN/m2
"o = Yw (hw +hc- h'} =9.81(3 + 4 - 1.2) = 57 kN/m2

410 Chapter 10
4:1 3:1 2:1
0.12
0.10
0.08
?L
^0.06
j
0.04
0.02
4:1 3:1 2:1 1.5:
2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34
Slope angle/?, degrees
Figure 10.30 Stability charts (after Spencer, 1967)

Figure Ex. 10.15
Therefore tf0 =(JQ - UQ = 113 - 57 = 56 kN/m2
After drawdown,
o= ysathc = 21 x 4 = 84 kN/m2
u = yw (hc - h'} =9.81(4 - 1.2) = 27.5 kN/m2
of
=a-u =S4-27.5 = 56.5 kN/m2
The change in strength is zero since the effective vertical stress does not change.
Note: There is no change in strength due to sudden drawdown but the direction of forces of
the seepage water changes from an inward direction before drawdownto an outward direction after
drawdown and this is the main cause for the reduction in stability.
10.18 PROBLEMS
10.1 Find the critical height of an infinite slope having a slope angle of 30°. The slope is made of
stiff clay having a cohesion 20 kN/m2
, angle of internal friction 20°, void ratio 0.7 and
specific gravity 2.7. Consider the following cases for the analysis.
(a) the soil is dry.
(b) the water seeps parallel to the surface of the slope.
(c) the slope is submerged.
10.2 An infinite slope has an inclination of 26° with the horizontal. It is underlain by a firm
cohesive soil having Gs = 2.72 and e = 0.52. There is a thin weak layer 20 ft below and
parallel to the slope (c' - 525 lb/ft2
, 0' = 16°). Compute the factors of safety when (a) the
slope is dry, and (b) ground water flows parallel to the slope at the slope level.
10.3 An infinite slope is underlain with an overconsolidated clay having c' - 210 lb/ft2
, 0' = 8°
and ysat = 120 lb/ft3
. The slope is inclined at an angle of 10° to the horizontal. Seepage is
parallel to the surface and the ground water coincides with the surface. If the slope fails
parallel to the surface along a plane at a depth of 12ft below the slope, determine the factor
of safety.
10.4 A deep cut of 10 m depth is made in sandy clay for a road. The sides of the cut make an
angle of 60° with the horizontal. The shear strength parameters of the soil are
c' - 20 kN/m2
, fi =25°, and 7= 18.5 kN/m3
. If AC is the failure plane (Fig Prob. 10.4),
estimate the factor of safety of the slope.

412 Chapter 10
y= 18.5kN/m3
Figure Prob. 10.4
W = 1050 kN
Figure Prob. 10.5
10.5 A 40° slope is excavated to a depth of 8 m in a deep layer of saturated clay having strength
parameters c = 60 kN/m2
, 0 = 0, and y= 19 kN/m3
. Determine the factor of safety for the
trial failure surface shown in Fig. Prob. 10.5.
10.6 An excavation to a depth of 8 m with a slope of 1:1 was made in a deep layer of saturated
clay having cu = 65 kN/m2
and 0M = 0. Determine the factor of safety for a trial slip circle
passing through the toe of the cut and having a center as shown in Fig. Prob. 10.6. The unit
weight of the saturated clay is 19 kN/m3
. No tension crack correction is required.
10.7 A 45° cut was made in a clayey silt with c = 15 kN/m2
, 0 = 0 and y = 19.5 kN/m3
. Site
exploration revealed the presence of a soft clay stratum of 2 m thick having c = 25 kN/m2
and 0 = 0 as shown in Fig. Prob. 10.7. Estimate the factor of safety of the slope for the
assumed failure surface.
10.8 A cut was made in a homogeneous clay soil to a depth of 8 m as shown in Fig. Prob. 10.8. The
total unit weight of the soil is 18 kN/m3
, and its cohesive strength is 25 kN/m2
.

<§)
Figure Prob. 10.6
Figure Prob. 10.7
Assuming a 0 = 0 condition, determine the factor of safety withrespect to a slip circle passing
through the toe. Consider a tension crack at the end of the slip circle on the top of the cut.
10.9 A deep cut of 10 m depth is made in natural soil for the construction of a road. The soil
parameters are: c' = 35 kN/m2
, 0' = 15° and 7= 20 kN/m3
.
Figure Prob. 10.8

414 Chapter 10
Figure Prob. 10.9
The sides of the cut make angles of 45° with the horizontal. Compute the factor of safety
using friction circle method for the failure surface AC shown in Fig. Prob. 10.9.
10.10 An embankment is to be built to a height of 50 ft at an angle of 20° with the horizontal. The
soil parameters are: c' - 630 lb/ft2
, 0' = 18°and 7= 115 lb/ft3
.
Estimate thefollowing;
1. Factor of safety of the slope assuming full friction is mobilized.
2. Factor of safety with respect to friction if the factor of safety with respect to cohesion is
1.5.
Use Taylor's stability chart.
10.11 A cut was made in natural soil for the construction of a railway line. The soil parameters
are: c' = 700 lb/ft2
, 0' = 20° and 7= 110 lb/ft3
.
Determine the critical height of the cut for a slope of 30° with the horizontal by making use
of Taylor's stability chart.
10.12 An embankment is to be constructed by making use of sandy clay having the following
properties: c' = 35 kN/m2
, 0' = 25° and y= 19.5 kN/m3
.
The height of the embankment is 20 m with a slope of 30° with the horizontal as shown in
Fig. Prob. 10.12. Estimate the factor of safety by the method of slices for the trial circle
shown in the figure.
10.13 If an embankment of 10 m height is to be made from a soil having c' =25 kN/m2
, 0' = 15°,
and 7=18 kN/m3
, what will be the safe angle of slope for a factor of safety of 1.5?
10.14 An embarkment is constructed for an earth dam of 80 ft high at a slope of 3:1. The
properties of the soil used for the construction are: c - 770 lb/ft2
, 0' = 30°, and
7=110 lb/ft3
. The estimated pore pressuer ratio r =0.5. Determine the factor of safety by
Bishop and Morgenstern method.
10.15 For the Prob. 10.14, estimate the factor of safety for 0'= 20°. All the other data remain the
same.
10.16 For the Prob. 10.14, estimate the factor of safety for a slope of 2:1 with all the oother data
remain the same.

Figure Prob. 10.12
10.17 A cut of 25 m dopth is made in a compacted fill having shear strength parameters of
c = 25 kN/m2
, and 0' = 20°. The total unit weight of the material is 19 kN/m3
. The pore
pressuer ratio has an average value of 0.3. The slope of the sides is 3:1. Estimate the factor
of safety using the Bishop and Morgenstern method.
10.18 For the Prob. 10.17, estimate the factor of safety for 0'= 30°, with all the other data remain
the same.
10.19 For the Prob. 10.17, esatimate the factor of safety for a slope of 2:1 with all the other data
remaining the same.
10.20 Estimate the minimum factor of safety for a complete drawdown condition for the section
of dam in Fig. Prob. 10.20. The full reservoir level of 15 m depth is reduced to zero after
drawdown.
10.21 What is the safety factor if the reservoir level is brought down from 15 m to 5 m depth in
the Prob. 10.20?
10.22 An earth dam to be constructed at a site has the following soil parameters: c'= 600 lb/ft2
,
y = 110 lb/ft3
, and 0'= 20°. The height of of dam H = 50 ft.
The pore pressure ratio ru =0.5. Determine the slope of the dam for a factor of safety of 1.5
using Spencer's method (1967).
c' = 15 kN/m2
<f>' = 30°
y = 20 kN/m3
Figure Prob. 10.20

416 Chapter 10
O
R = 45 ft
15ft
Figure Prob. 10.24
10.23 If the given pore pressure ratio is 0.25 in Prob. 10.22, what will be the slope of the dam?
10.24 An embankment has a slope of 1.5 horizontal to 1 vertical with a height of 25 feet. The soil
parameters are:
c - 600 lb/ft2
, 0' = 20°, and 7= 110 lb/ft3
.
Determine the factor of safety using friction circle method for the failure surface AC shown
in Fig. Prob. 10.24.
10.25 It is required to construct an embankment for a reservoir to a height of 20 m at a slope of
2 horizontal to 1 vertical. The soil parameters are:
c = 40 kN/m2
, f = 18°, and 7= 17.5 kN/m3
.
Estimate the following:
1. Factor of safety of the slope assuming full friction is mobilized.
2. Factor of safety with respect to friction if the factor of safety with respect to cohesion is
1.5.
Use Taylor's stabilitychart.
10.26 A cutting of 40 ft depth is to be made for a road as shown in Fig. Prob. 10.26. The soil
properties are:
c' = 500 lb/ft2
, 0' = 15°, and 7= 115 lb/ft3
.
Estimate the factor of safety by the method of slices for the trial circle shown in the figure.
10.27 An earth dam is to be constructed for a reservior. The height of the dam is 60 ft. The
properties of the soil used in the construction are:
c = 400 lb/ft2
, 0° = 20°, and 7= 115 lb/ft3
, and ft = 2:1.
Estimate the minimumfactor of safety for the complete drawn from the full reservior level
as shown in Fig. Prob. 10.27 by Morgenstern method.
10.28 What is the factor of safety if the water level is brought down from 60 ft to 20 ft above the
bed level of reservoir in Prob. 10.27?

c' = 5001b/ft2
0'=15°
y=1151b/ft3
Figure Prob. 10.26
Full reservoir level
1
Figure Prob. 10.27
10.29 For the dam given in Prob. 10.27, determine the factor of safety for r«= 0.5 by Spencer's
method.

Chapter 10

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